oldhw2solution_pdf

oldhw2solution_pdf - klarin (sjk772) – oldhomework 02 –...

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Unformatted text preview: klarin (sjk772) – oldhomework 02 – Turner – (56705) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Three charges are arranged as shown in the figure. x y 4 . 9 nC 4 . 4 m − . 5 nC . 88 nC − 1 . 5 m Find the magnitude of the electro- static force on the charge at the origin. The Coulomb cosnstant is 8 . 98755 × 10 9 N · m 2 / C 2 . Correct answer: 2 . 09348 nN. Explanation: Let : ( x ,y ) = (0 m , 0 m) , q = − . 5 nC , ( x 1 ,y 1 ) = (4 . 4 m , 0 m) , q 1 = 4 . 9 nC , ( x 2 ,y 2 ) = (0 m , − 1 . 5 m) , and q 2 = 0 . 88 nC . x y q 1 x 1 q q 2 y 2 F θ The force vector F 10 (in the hatwider 10 direction, − ˆ ı ) be- tween charge − 5 × 10 − 10 C and 4 . 9 × 10 − 9 C is vector F 10 = + k e q q 1 x 2 1 ( − ˆ ı ) = ( − 8 . 98755 × 10 9 N · m 2 / C 2 ) × ( − . 5 nC) (4 . 9 nC) (4 . 4 m) 2 ˆ ı = (1 . 13737 × 10 − 9 N)ˆ ı. The plus sign means the force vector F 10 is towards the positive x-axis. The force vector F 20 (in the hatwider 20 direction, +ˆ ) between charge − 5 × 10 − 10 C and 8 . 8 × 10 − 10 C is vector F 20 = + k e q q 2 y 2 2 (+ˆ ) = (8 . 98755 × 10 9 N · m 2 / C 2 ) × ( − . 5 nC) (0 . 88 nC) ( − 1 . 5 m) 2 ˆ = ( − 1 . 75757 × 10 − 9 N)ˆ . The minus sign means the force vector F 20 is towards the negative y-axis. The magni- tude of the total force exerted on the charge − 5 × 10 − 10 C is equal to bardbl vector F bardbl = radicalBig F 2 10 + F 2 20 = bracketleftBig (1 . 13737 × 10 − 9 N) 2 + ( − 1 . 75757 × 10 − 9 N) 2 bracketrightBig 1 2 = 2 . 09348 × 10 − 9 N = 2 . 09348 nN . 002 (part 2 of 2) 10.0 points What is the angle θ between the electrostatic force on the charge at the origin and the pos- itive x-axis? Answer in degrees as an angle between − 180 ◦ and 180 ◦ measured from the positive x-axis, with counterclockwise posi- tive. Correct answer: − 57 . 0919 ◦ . Explanation: The angle between vector F and the negative y- axis is θ = arctan parenleftbigg − 1 . 75757 × 10 − 9 N 1 . 13737 × 10 − 9 N parenrightbigg = − 57 . 0919 ◦ . klarin (sjk772) – oldhomework 02 – Turner – (56705) 2 003 10.0 points Two identical small charged spheres hang in equilibrium with equal masses as shown in...
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This note was uploaded on 10/10/2010 for the course PHY 56705 taught by Professor Turner during the Fall '10 term at University of Texas.

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oldhw2solution_pdf - klarin (sjk772) – oldhomework 02 –...

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